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KIM [24]
4 years ago
6

Sodium hydroxide, NaOH, is a strong base that is used in industrial synthesis and processes such as making paper. What is the ma

ss of 2.60 ×1022 molecules of NaOH (Molar mass = 40.0 g/mol)?
Chemistry
2 answers:
Ghella [55]4 years ago
7 0
Answer : Mass would be = 1.859 g for NaOH

Explanation : 
We have got 2.88 X  10^{22} molecules of NaOH,

so we get,

\frac{2.88 X 10^{22} moles of NaOH }{6.022 X 10^{23}/mol } X 40.0 g of NaOH/mol

On solving we get, answer as 1.859 g

Hence the g of NaOH will be 1.86.
elena-14-01-66 [18.8K]4 years ago
3 0
  The  mass of 2.60 x10^22  molecules of NaOH  is   1.728   grams

calculation
 mass = moles  x molar mass
   molar mass = 40 g/mol

mole   is calculated   using the Avogadro   law
that is  1 mole =   6.02 x10^23 molecules
              ?        =  2.60  x10 ^ 22  molecules

by   cross   multiplication

 (1 mole x  2.60  x10^22) /  6.02 x0^23  =  0.0432  moles

mass   is therefore= 0.0432  mole x  40 g/mol = 1.728 grams
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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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